I'm trying to learn the epsilon delta proof.
Given
$$f(x,y) = \frac{x^2|y|}{x^2+y^2} \text{ for } (x,y) \neq (0,0)$$
I want an epsilon delta proof to see that $f$ is continuous.
I have seen this proof on the internet for $\frac{x^2y}{x^2+y^2}$ , but I don't quite understand it.
Let $\epsilon>0$. Take $\delta:=\epsilon$
If $\vert \textbf{x} \vert < \delta$, then $|y| < \delta$ (where $\textbf{x}=(x,y)$). Therefore:
$\displaystyle |f(x,y)| =\frac{|x^2y|}{|x^2+y^2|}< |y|<\epsilon$
As far as I know, an $(\epsilon, \delta)$ proof is structured like this:
Step $0$: Let $\epsilon > 0$
Step $1$: We try to simplify $|f(x)-f(x_0)|$
Step $2$: We try to find a $\delta$, so that $|f(x)-f(x_0)|$ is $< \epsilon$
Step $3$: Summarize: Let $\epsilon > 0$. With $\delta$ we can follow, that for all $x$ it holds that $|f(x) - f(x_0)| < \delta \leq \epsilon$
I don't see how that above is a full proof.
What about $|f(x)-f(x_0)|$? Why do we only say $|f(x,y)| =\frac{|x^2y|}{|x^2+y^2|}$ and how do we know what $\delta$ we should take? (In this case he took $\delta := \epsilon)$ Could we have used $\delta = \frac{1}{2} \epsilon$ instead?
I think something is missing when defining the function. i.e.
$f(x,y)=$ $\begin{cases}\frac{x^2|y|}{x^2+y^2} \text{ for } (x,y) \neq (0,0) \\0 \text{ for } (x,y) = (0,0) \end{cases}$.
Then you can proceed.
Let $\epsilon\gt0$.
Observe that $|f(x,y)-f(0,0)|=|f(x,y)|=\frac{|x^2||y|}{|x^2+y^2|}\lt \frac{|x^2||y|}{|x^2|}\lt|y|$
Let $\epsilon=\delta$.
Suppose $|(x,y)-(0,0)|=|\sqrt{x^2+y^2}|\lt\delta \implies |x|,|y|\lt\delta$.
So $|f(x,y)-f(0,0)|=\frac{|x^2||y|}{|x^2+y^2|}\lt |y|\lt\delta=\epsilon $.
for the part can't I use $\delta=\frac{1}{2}\epsilon$. Yes of course you can. In fact you can use any $\delta\lt\epsilon$.